Problem: The equation of an ellipse is given below. $\dfrac{(x+6)^2}{1}+\dfrac{(y+8)^2}{49}=1$ What is its center ? $($
Standard equation of an ellipse The standard equation of an ellipse with a center $ ( h, k)$ is given below. $\dfrac{(x{-h})^2}{a^2}+\dfrac{(y{-k})^2}{b^2}=1$ In this equation, $a$ represents the radius in the $x$ -direction (horizontal radius), and $b$ represents the radius in the $y$ -direction (vertical radius). The major radius is the longer of the two, and the minor radius is the shorter of the two. Writing the given equation in standard form We can rewrite the given equation in standard form to reveal the center and radii of the ellipse. $\dfrac{(x-{(-6)})^2}{1^2}+\dfrac{(y-{(-8)})^2}{7^2}=1$ From the equation, we can see that the ellipse is centered at $( {-6}, {-8})$, $a=1$, and $b=7$. Since $b>a$, it follows that $b$ is the major radius, and $a$ is the minor radius. This means that the ellipse is taller than it is wide. Summary The ellipse is centered at $ (-6,-8)$. It has a major radius of $7$ units. It has a minor radius of $1$ unit.